| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Find alpha from CI width |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring standard formulas. Part (a) is trivial (midpoint), part (b) uses the CI width formula with known critical value, part (c) reverses this to find a critical value, and part (d) applies the sample size formula. All parts are routine applications of S3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{26.624 + 28.976}{2} = 27.8\) | B1 | For 27.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(28.976 - 26.624 = 2 \times 1.96 \times \dfrac{\sigma}{\sqrt{25}}\) or \(26.624 = 27.8 - 1.96 \times \dfrac{\sigma}{\sqrt{25}}\) or \(28.976 = 27.8 + 1.96 \times \dfrac{\sigma}{\sqrt{25}}\) | M1 B1 | M1: correct structure; B1: awrt 1.96; where \(1.5 < z < 2.4\) |
| \(\sigma = 3\) | A1* cso | Answer is given so no incorrect working must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2 \times z \times \dfrac{3}{\sqrt{25}} = 2.1\), so \(z = 1.75\) | M1 A1 | M1: for \(2 \times z \times \dfrac{3}{\sqrt{25}} = 2.1\); A1: \(z = 1.75\) |
| \(P(Z > 1.75) = P(Z < -1.75) = 1 - 0.9599 = 0.0401\) | M1 A1ft | M1: for \(1-p\) where \(p\) is a probability; A1ft: 0.0401 or ft their \(z\) value (allow 0.04) |
| Confidence level \(= 100 \times (1 - 2 \times 0.0401) = 91.98\%\) | M1 A1 | M1: for \(100\times(1 - 2\times 0.0401)\) ft their \(P(Z<-1.75)\); A1: awrt 92.0 (allow 92) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2 \times 1.96 \times \dfrac{3}{\sqrt{n}} < 1.5\) | M1 | For \(2\times z \times \dfrac{3}{\sqrt{n}} < 1.5\); \(z\) must be correct or consistent with (b); allow \(\leq\) or \(\geq\) |
| \(\sqrt{n} > \dfrac{6 \times 1.96}{1.5}\) | dM1 | Dependent on previous M; correct rearrangement to get \(\sqrt{n} > \ldots\) or \(n > \ldots\); allow \(\geq\) |
| \(\sqrt{n} >\) awrt \(7.84\), so \(n = 62\) | A1 A1 | A1: awrt 7.84 (may be implied by awrt 61.5); A1: \(n = 62\) |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{26.624 + 28.976}{2} = 27.8$ | B1 | For 27.8 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $28.976 - 26.624 = 2 \times 1.96 \times \dfrac{\sigma}{\sqrt{25}}$ or $26.624 = 27.8 - 1.96 \times \dfrac{\sigma}{\sqrt{25}}$ or $28.976 = 27.8 + 1.96 \times \dfrac{\sigma}{\sqrt{25}}$ | M1 B1 | M1: correct structure; B1: awrt 1.96; where $1.5 < z < 2.4$ |
| $\sigma = 3$ | A1* cso | Answer is given so no incorrect working must be seen |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2 \times z \times \dfrac{3}{\sqrt{25}} = 2.1$, so $z = 1.75$ | M1 A1 | M1: for $2 \times z \times \dfrac{3}{\sqrt{25}} = 2.1$; A1: $z = 1.75$ |
| $P(Z > 1.75) = P(Z < -1.75) = 1 - 0.9599 = 0.0401$ | M1 A1ft | M1: for $1-p$ where $p$ is a probability; A1ft: 0.0401 or ft their $z$ value (allow 0.04) |
| Confidence level $= 100 \times (1 - 2 \times 0.0401) = 91.98\%$ | M1 A1 | M1: for $100\times(1 - 2\times 0.0401)$ ft their $P(Z<-1.75)$; A1: awrt 92.0 (allow 92) |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2 \times 1.96 \times \dfrac{3}{\sqrt{n}} < 1.5$ | M1 | For $2\times z \times \dfrac{3}{\sqrt{n}} < 1.5$; $z$ must be correct or consistent with (b); allow $\leq$ or $\geq$ |
| $\sqrt{n} > \dfrac{6 \times 1.96}{1.5}$ | dM1 | Dependent on previous M; correct rearrangement to get $\sqrt{n} > \ldots$ or $n > \ldots$; allow $\geq$ |
| $\sqrt{n} >$ awrt $7.84$, so $n = 62$ | A1 A1 | A1: awrt 7.84 (may be implied by awrt 61.5); A1: $n = 62$ |
---\begin{enumerate}
\item The random variable $X$ is normally distributed with unknown mean $\mu$ and known variance $\sigma ^ { 2 }$
\end{enumerate}
A random sample of 25 observations of $X$ produced a $95 \%$ confidence interval for $\mu$ of (26.624, 28.976)\\
(a) Find the mean of the sample.\\
(b) Show that the standard deviation is 3
The $a$ \% confidence interval using the 25 observations has a width of 2.1\\
(c) Calculate the value of $a$\\
(d) Find the smallest sample size, of observations from $X$, that would be required to obtain a 95\% confidence interval of width at most 1.5
\hfill \mbox{\textit{Edexcel S3 2022 Q3 [14]}}